Property of Maximal function

Question

Hardy littlewood Maximal function is defined for for a locally integrable function on $\mathbb{R}^n$ by $$Mf(x) =\sup \frac{1}{|Q|} \int_Q |f(y)|dy$$ where supremum is taken over all cubes Q in $\mathbb{R}^n$ containing x (Book "interpolation of operators"). The book says that it is easy to see that $$Mf(x) \geq c/|x|^n, ~ (|x| \geq 1).$$ I don't see how to see the fact. While searching online, I got one clue that there exists a cube Q such that $\int_Q |f|dy>0$. But still I can't see how to prove. May be i am missing some trivial thing. Please help to figure it out

Answer 1

Denote by $Q_a(x)$ the cube centered at $x$ with side $a$.

Assume, without loss of generality, that $f\geq0$. Suppose that $f$ is not zero a.e. Then there exists some cube $Q_r(x_0)$ with $c_0=\int_{Q_r(x_0)}f>0$. Any point in $Q_r(x_0)$ is at most at distance $s=\sqrt n\,(\|x_0\|+r)$ from the origin.

For any $|x|\geq1$ we have $|sx|\geq s$, which implies $Q_r(x_0)\subset Q_{s\|x\|}(x)$. Then $$ Mf(x)\geq\frac1{m(Q_{s\|x\|})}\,\int_{Q_{s\|x\|}(x)}f \geq\frac1{m(Q_{s\|x\|})}\,\int_{Q_{r}(x_0)}f \geq\frac {c_0}{m(Q_{s\|x\|})}=\frac{c_0/s^n}{\|x\|^n}. $$